PRS Combination Exam 3

W11D2 Concept Questions

A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational moment of inertia and her angular speed increases. Assume that her angular momentum is constant. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be

  1. the same.
  2. larger.
  3. smaller.

Answer: 2. When she pulls her arms in there are no external torques so her angular momentum is constant. The magnitude of her angular momentum about the rotation axis passing through the center of the stool is . Her kinetic energy is

.

Since is constant and her moment of inertia decreases when she pulls her arms, her kinetic energy must increase.

Concept Question: A tetherball of mass m is attached to a post of radius by a string. Initially it is a distance from the center of the post and it is moving tangentially with a speed . The string passes through a hole in the center of the post at the top. The string is gradually shortened by drawing it through the hole. Ignore gravity. Until the ball hits the post,

  1. The energy and angular momentum about the center of the post are constant.
  1. The energy of the ball is constant but the angular momentum about the center of the post changes.
  1. Both the energy and the angular momentum about the center of the post, change.
  1. The energy of the ball changes but the angular momentum about the center of the post is constant.

Answer: 4.The tension force points radially in; so torque about central point is zero: angular momentum about central point is constant.

For all radial forces, (example gravitation), the angular momentum about the central

point is a constant of the motion.

A small displacement of the ball has a radially component inward so the work done by tension is not zero.,

.

Hence the mechanical energy is not constant.

Concept Question: A tetherball of mass m is attached to a post of radius by a string. Initially it is a distance from the center of the post and it is moving tangentially with a speed . The string wraps around the outside of the post. Ignore gravity. Until the ball hits the post,

  1. The energy and angular momentum about the center of the post are constant.
  1. The energy of the ball is constant but the angular momentum about the center of the post changes.
  1. Both the energy of the ball and the angular momentum about the center of the post, change.
  1. The energy of the ball changes but the angular momentum about the center of the post is constant.

Answer 2. The tension force points towards contact point; so torque about central point is not zero: angular momentum about central point is not constant.

Small displacement is always perpendicular to string since at each instant in time ball undergoes instantaneous circular motion about string contact point with pole.

Therefore the tension force is perpendicular to the displacement

.

Hence mechanical energy is constant.

Concept Question: A streetcar is freely coasting (no friction) around a large circular track. It is then switched to a small circular track.

When coasting on the smaller circle the streetcar's

  1. mechanical energy is conserved and angular momentum about the center is constant
  2. mechanical energy is not conserved and angular momentum about the center is constant
  3. mechanical energy is not conserved and angular momentum about the center is not constant
  4. mechanical energy is conserved and angular momentum about the center is not constant.

Answer: 4. Carefully draw a free body diagram for the streetcar while it is on the link of track connecting the two circular tracks. In the vertical direction, the force of gravity at all times balances the vertical component of the normal force of the track on the car, so there is no acceleration in the vertical direction and we may ignore it for the rest of this problem. Along the plane of motion, the wheels are guided by component of the normal force in the plane. This force is always normal to the direction of motion, so no work is done by this force. Therefore the streetcar's total energy does not change, and its speed remains constant. Note that this force does exert a torque around the center of the circles because it does not point radially inward, and therefore its angular momentum is not constant!

W13D1 Concept Questions

Concept Question: Two disks are separated by a spindle of smaller diameter. A string is wound around the spindle and pulled gently. Which positions of the string cause the assembly to roll to the right?

1) Only A

2) Only B

3) Only C

4) A and B

5) B and C

6) A and B and C

7) None of the configurations shown

Answer: 1. When the string is pulled in direction A, the torque due to the pulling force is out of the page and if there were no friction between the surfaces, the assembly would spin counterclockwise and slide to the right. If there is some contact friction between the surfaces, the torque due to the friction is into the page which will cause a clockwise rotation. For small pulling forces, the frictional torque can exceed the pulling torque and the assembly will roll to the right.

The force equation in the vertical direction yields so the normal force decreases as the angle increases, according to . Therefore the kinetic fiction force decreases (since ) and can no longer supply enough torque to keep the wheel spinning clockwise and rolling to the right. As the direction of the pulling force is shifted towards B, there will be a critical angle in which the yo-yo will stop rotating, and the yo-yo will slide to the right with kinetic friction opposing the motion. At any larger angle the yo-yo rotates counterclockwise and so cannot roll to the right.

Concept Question: A physical pendulum consists of a uniform rod of length and mass m pivoted at one end. A disk of mass and radius is fixed to the other end. Suppose the disk is now mounted to the rod by a frictionless bearing so that is perfectly free to spin. Does the period of the pendulum

  1. increase?
  2. stay the same?
  3. decrease?

Answer: 3. When the disk is fixed to the end, the pendulum is a rigid body. As the pendulum swings, the disk is rotating about its center of mass. Thus some potential energy is transformed into rotational kinetic energy and the rest is transformed into translational kinetic energy of the center of mass. When the disk is mounted on a frictionless bearing it will not rotate about its center of mass as the pendulum swings. Hence all of the gravitational potential energy goes into translational kinetic energy of the center of mass, and so the disk will move faster. Thus it will take less time to complete one swing and hence the period decreases.

Concept Question: Two cylinders of the same size and mass roll without slipping down an incline, starting from rest. Cylinder A has most of its mass concentrated at the rim, while cylinder B has most of its mass concentrated at the center. Which has more total kinetic energy at the bottom?

  1. A
  2. B
  3. Both have the same

Answer: 3. Since the cylinders start form the same height, the same amount of gravitational potential energy is converted to kinetic energy. Therefore they have the same kinetic energy at the bottom.

Concept Question: Two cylinders of the same size and mass roll without slipping down an incline, starting from rest. Cylinder A has most of its mass concentrated at the rim, while cylinder B has most of its mass concentrated at the center. Which reaches the bottom first?

  1. A
  2. B
  3. Both have the same

Answer: 2. The kinetic energy of a cylinder is given by

.

Since the cylinder is rolling without slipping, if is the radius of the cylinders, then . So the kinetic energy becomes

.

Since the cylinders start form the same height, the same amount of gravitational potential energy is converted to kinetic energy. Since cylinder A has its mass concentrated at the rim, it has larger moment of inertia than cylinder B. Therefore a larger moment of inertia means a smaller velocity at the bottom. So for the same drop in height, cylinder B has a faster velocity hence it will reach the bottom first.

W13D2 Concept Questions:

Concept Question: What impulse must be given to the ball in order to rotate its orbit by 90 degrees as shown without changing its speed?

Answer: h. must halt the y motion and provide a momentum of equal magnitude along the z direction.

cancels the z component of and adds a component of the same magnitude in the negative y direction.

Concept Question: To make the top of the shaft move in the -y direction in which direction should one apply the top half of an impulse couple?

Answer: a. The impulse couple Jb applied to the shaft has the same effect as the Ja couple applied directly to the masses. Both produce a torque in the - y direction.

W14D2 Concept Questions

Concept Question: The angular momentum about the point O of the “reduced body”

  1. is constant.
  2. changes throughout the motion because the speed changes.
  3. changes throughout the motion because the distance from O changes.
  4. changes throughout the motion because the angle  changes.
  5. Not enough information to decide.
  6. We had the angular momentum quiz last Friday so I don’t need to think about it anymore.

Answer: 1. The torque about the point O due to the central force is zero,

,

since the vector and the vector are anti-parallel. Therefore the change in angular momentum is zero

.

For motion confined to the plane, the angular momentum must be constant.

Concept Question: The mechanical energy of the “reduced body”

  1. is constant.
  2. changes throughout the motion because the speed changes.
  3. changes throughout the motion because the distance from O changes.
  4. is not constant because the orbit is not zero hence the central force does work.
  5. Not enough information to decide.

Answer: 1. There are no non-conservative forces so the mechanical energy is constant.

W15D1 Concept Questions

Concept Question:The angular momentum of the planet is conserved about:

  1. the center of the ellipse
  2. the focus of the ellipse at the sun
  3. the focus of the ellipse opposite the sun
  4. all three above
  5. two of the three above
  6. none of the above

Answer: 2. The central point of the planet’s orbit is the focus of the ellipse at the sun., and hence the point where the gravitational force is directed.

The torque about that focus pointdue to the central force is zero,

,

since the vector and the vector are anti-parallel. Therefore the angular momentum must be constant about the focus of the ellipse at the sun.